168
7 Quadrupole Contributions from Interface and Bulk
In order that Eq. (7.49) is equivalent to (7.51), these two elements should be
converted to
χ
B
G,yyz ((, ω 1 , ω 2 )=l G ζ
Q1,β
2
((, ω 1 , ω 2 )
k
β
T (ω 1 )×k
β
T (ω 2 )
y
k x (ω 2 )
f
β
y (()f
β
y (ω 1 )f
β
z (ω 2 ).
(7.54)
[Problem 7.2] Derive χ B
G,yyz in Eq. (7.54) from χ B0
G in Eqs. (7.52) and (7.53).
(Hint) During the derivation, use the following two relations,
cos θ α
I (ω 2 )
sin θ α
I (ω 2 )
=
q α (ω 2 )
k x (ω 2 )
,
f
β
x
f
β
z
= ε
β ,
and L I (ω f ) in Eq. (5.30).
Analogous reformulation is possible for other polarization combinations. The
relevant tensor elements of χ B
G in Eq. (7.50) to the four polarization combinations
are summarized as follows.
• SSP case:
χ
B
G,yyz ((, ω 1 , ω 2 ) =
l G ζ
Q1,β
2
((, ω 1 , ω 2 )
k
β
T (ω 1 ) × k
β
T (ω 2 )
y
k x (ω 2 )
f
β
y (()f
β
y (ω 1 )f
β
z (ω 2 ),
(7.54)
• SPS case:
χ
B
G,yzy ((, ω 1 , ω 2 ) =
l G ζ
Q2,β
1
((, ω 1 , ω 2 )
k
β
T (ω 2 ) × k
β
T (ω 1 )
y
k x (ω 1 )
f
β
y (()f
β
z (ω 1 )f
β
y (ω 2 ),
(7.55)
• PSS case:
χ
B
R,zyy ((, ω 1 , ω 2 ) = l R
1
k x (()
ζ
Q1,β
3
((, ω 1 , ω 2 )
k
β
T (ω 1 ) × k
β
R (()
y
+ζ
Q2,β
3
((, ω 1 , ω 2 )
k
β
T (ω 2 ) × k
β
R (()
y
f
β
z (()f
β
y (ω 1 )f
β
y (ω 2 ),
(7.56)
χ
B
T ,zyy ((, ω 1 , ω 2 ) =